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\section*{1}
result = $1.11011101 \times 2^8$.
\section*{2}
result = $1.001001001\cdots \times 2^{-1}$.
\section*{3}
$$x = 1.000\cdots0 \times \beta^e$$
$$x_L = (\beta-1).(\beta-1)(\beta-1)\cdots(\beta-1) \times \beta^{e-1}$$
$$x_R = 1.000\cdots1 \times \beta^e$$
$$x_R - x = \beta ^{e-p+1}=\beta \beta ^{e-p} = \beta (x - x_L)$$
\section*{4}
$$x_L = 1.00100100100100100100100 \times 2^{-1}$$  
$$x_R = 1.00100100100100100100101 \times 2^{-1}$$
$$fl(x) = x_R$$
$$re(x) = 0.4 \times 2^{-24}$$
\section*{5}
$$\epsilon_u = \frac{1}{2} \epsilon_M = 2^{-24}$$
\section*{6}
$$\cos \frac{1}{4} = 1.111100000001\cdots \times 2^{-1}$$
$$1 = 1.00000\cdots \times 2^0$$
By theorem 4.49,
$$2^{-6} < 1- \cos \frac{1}{4} < 2^{-5}$$
There are 5 to 6 bits of precision lost in the subtraction.
\section*{7}
\subsection*{(1)}
According to Taylor Formula,
$$1-\cos(x) = \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} -\cdots$$
\subsection*{(2)}
$$1-\cos(x) = 2\sin^2(\frac{x}{2})$$
\section*{8}
According to Definition 4.59,
$$C_f(x) = |\frac{x f'(x)}{f(x)}|$$
\subsection*{(1)}
$$C_f(x) = |\frac{\alpha x}{x-1}|$$
$C_f(x)$ is large when $x \rightarrow 1$.
\subsection*{(2)}
$$C_f(x) = |\frac{1}{\ln(x)}|$$
$C_f(x)$ is large when $x \rightarrow 1$.
\subsection*{(3)}
$$C_f(x) = |x|$$
$C_f(x)$ is large when $x \rightarrow \pm \infty$.
\subsection*{(4)}
$$C_f(x) = |\frac{x}{\arccos(x) \sqrt{1-x^2}}|$$
$C_f(x)$ is large when $x \rightarrow \pm 1$.
\section*{9}
\subsection*{(1)}
$$\forall x \in [0, 1],C_f(x) = |\frac{x}{e^x-1}| \leq 1 $$
\subsection*{(2)}
$$f_{A}(x)=1-(1+\delta_1)e^{-(1+\delta_2)x}=1-e^{x_{A}}$$
$$x_{A}=(1+\delta_2)x-\ln(1+\delta_1)$$
$$cond_{A}(x)=\frac{1}{\epsilon_u}\inf\limits_{f(x_A)=f_A(x)} \frac{|x_A-x|}{|x|}$$
$$cond_{A}(x)=\frac{1}{\epsilon_u}\inf(\delta_2-\ln(1+\delta_1))\leq 1+\frac{1}{x}$$
\subsection*{(3)}
\includegraphics[scale=0.5]{9.png}

From the graph we can see that $cond_f$ is small on the whole interval and $cond_A$ is very large when $x \to 0$.
\section*{10}
$$cond_{f}(a)=||A(x)||_{1}=\sum_{i=0}^{n-1}|\frac{a_{i}\frac{\partial{r}}{\partial{a_{i}}}}{r}|$$
$$q'(r)\frac{\partial{r}}{\partial{a_{i}}}+r^i=0$$
$$\frac{\partial{r}}{\partial{a_{i}}}=-\frac{r^i}{q'(r)}$$
$$cond_{f}(x)=\frac{\sum_{i=0}^{n-1}|a_{i}r^i|}{|rq'(r)|}$$
In Wilkinson example,$q(x)=\prod_{i=1}^{n}(x-i)$ and $r_{n}=n$,
$cond_{f}(a)\geq \frac{n^n}{n!}$.
\section*{11}
Consider $p=2,a=1.0 \times 2^0,b=1.1 \times 2^0$.Then $\frac{a}{b}=\frac{2}{3}$.In the register of precision 4,$\frac{a}{b}=0.101 \times 2^0$.$fl(\frac{a}{b})=1.0 \times 2^{-1}$ and $E(\frac{a}{b})=0.01$.
It is contradictory.
\section*{12}
$\forall x \in [128,129]$,$x=a \times 2^7$,$a \in [1,\frac{129}{128}]$.
So the distance between two adjacent points is $2^7\epsilon_u=2^{-16}>10^6$.
Thus,the accuracy can't reach $10^{-6}$.

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